Just started on the topic Circuit Analysis, and am supposed to find the transfer function of the following circuit:
So I know I am supposed to use KVL to solve this problem.
I can first transform the circuit to:
From the circuit, I can see that because u*Vc is in parallel with the output:
$$V_o(s) = uV_c(s)$$
$$V_c(s)= \frac{V_o(s)}{u}$$
Using KVL,
$$V_i(s) = (I_1*R)+(I_3*R)+V_c(s)$$
where: $$V_c(s) = \frac{V_o(s)}{u}$$
Now the problem is, how do I exactly get I1 and I2 from the circuit?
New to this, thanks 🙂
EDIT: I crafted another KVL equation, as well as being able to determine I3.
$$V_i(s) = (I_1*R)+(I_2*1/sc)+V_0(s)$$
$$I_3(s) = \frac {V_c(s)}{1/sc}$$
Best Answer
First thing I did was redraw the circuit. I added a node 'Vx' to help me solve.
(1) $$I_1 = I_2 + I_3$$
(2) $$I_1 = \frac{V_i - V_x}{R}$$
(3) $$I_2 = (V_x - V_o) * sC$$
(4) $$I_3 = \frac{V_x - V_o/u}{R}$$
Plugging equations 2,3,4 into 1 gives:
(5) $$\frac{V_i - V_x}{R} = (V_x - V_o) sC +\frac{V_x - V_o/u}{R}$$
(6) $$V_i = V_x (sCR + 2) - V_o(sCR + 1/u)$$
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Now let's get Vx in terms of Vo. I do this using I_3
$$I_{3 (R)} = I_{3 (C)}$$
$$\frac{V_x - V_o/u}{R} = (V_o/u) * sC$$
$$V_x = V_o * \frac{sCR}{u} + \frac{V_o}{u}$$
$$V_x = V_o * \frac{sCR+1}{u}$$
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Now, I have the rest done, but believe that you are capable of taking it from here. I think your roadblock was using Vx, which helped me get I1, I2, and I3.
Check your work by setting 'u' equal to 3, you should see some major magnitude peaking. For picture below I used R = 10kΩ, C = 1uF, u = 3.